使用 LOOP 将单元格中的缺失值替换为上面 (n-1) 单元格中的值

Question

I have a data file with thousands of rows, that has gaps which I wish to fill with a value. I need to replace the empty cells with the values from those above it. It will be easier to give you an idea of what my data looks like, here is a sample

Variable <- c("AGE","","","","SEX","","SEGMENT","","","","")    
Value <- c(1, 2, 3, 4, 1, 2, 1, 2, 3, 4, 5)
Description <- c("18-24","25-34","35-44","45+","Female","Male","A","B","C","D","E")
df <- data.frame(Variable, Value, Description)

> df
   Variable Value Description
1       AGE     1       18-24
2               2       25-34
3               3       35-44
4               4         45+
5       SEX     1      Female
6               2        Male
7   SEGMENT     1           A
8               2           B
9               3           C
10              4           D
11              5           E

As you can see above the first column has gaps. I need these empty cells to be replaced with the relevant value above so the new variable will look like this in the dataframe

> df
   Variable Value Description Variable_NEW
1       AGE     1       18-24               AGE
2               2       25-34               AGE
3               3       35-44               AGE
4               4         45+               AGE
5       SEX     1      Female               SEX
6               2        Male               SEX
7   SEGMENT     1           A           SEGMENT
8               2           B           SEGMENT
9               3           C           SEGMENT
10              4           D           SEGMENT
11              5           E           SEGMENT

Thinking out aloud. I'm assuming to achieve this, I will need to create a new variable with a loop and then use a logic like this

    IF Variable[n]="" THEN Variable_New[n] = Variable[n-1], 
               ELSE Variable_New[n] = Variable[n]

I'm familiar with loops but don't how to write this kind of thing in R where it has a lag/n-1 kind of function. There are probably many ways to accomplish this, but it would be a preferable using a loop. Any help will be greatly appreciated. Thanks

Answer 1

Here a loop approach:

#Data
Variable <- c("AGE","","","","SEX","","SEGMENT","","","","")    
Value <- c(1, 2, 3, 4, 1, 2, 1, 2, 3, 4, 5)
Description <- c("18-24","25-34","35-44","45+","Female","Male","A","B","C","D","E")
df <- data.frame(Variable, Value, Description,stringsAsFactors = F)
#Create new column
df$NewVar <- df$Variable
#Loop
for(i in 2:dim(df)[1])
{
  df$NewVar[i] <- ifelse(df$NewVar[i]=="",df$NewVar[i-1],df$NewVar[i])
}

Output:

   Variable Value Description  NewVar
1       AGE     1       18-24     AGE
2               2       25-34     AGE
3               3       35-44     AGE
4               4         45+     AGE
5       SEX     1      Female     SEX
6               2        Male     SEX
7   SEGMENT     1           A SEGMENT
8               2           B SEGMENT
9               3           C SEGMENT
10              4           D SEGMENT
11              5           E SEGMENT

Answer 2

You don't need to write loops, there are built-in functions which can help you with this task.

You can replace blank values with NA and use fill :

library(dplyr)

df %>%
  mutate(Variable_NEW = replace(Variable, Variable == "", NA)) %>%
  tidyr::fill(Variable_NEW)

#   Variable Value Description Variable_NEW
#1       AGE     1       18-24          AGE
#2               2       25-34          AGE
#3               3       35-44          AGE
#4               4         45+          AGE
#5       SEX     1      Female          SEX
#6               2        Male          SEX
#7   SEGMENT     1           A      SEGMENT
#8               2           B      SEGMENT
#9               3           C      SEGMENT
#10              4           D      SEGMENT
#11              5           E      SEGMENT

Answer 3

You can write your own function with a loop or use the na.locf function from the zoo package to fill-in missing NA values. Example:

fillin <- function(x) {
  for (i in 2:length(x)) {
    if (x[i] %in% c(NA, "")) {
      x[i] <- x[i - 1]
    }
  }
  x
}

Variable <- c("AGE","","","","SEX","","SEGMENT","","","","")    
Value <- c(1, 2, 3, 4, 1, 2, 1, 2, 3, 4, 5)
Description <- c("18-24","25-34","35-44","45+","Female","Male","A","B","C","D","E")
df <- data.frame(Variable, Value, Description)

df$Variable_fillin <- fillin(df$Variable)

library(zoo)
df$Variable[df$Variable == ""] <- NA
df$Variable_nalocf <- na.locf(df$Variable)

df
#>    Variable Value Description Variable_fillin Variable_nalocf
#> 1       AGE     1       18-24             AGE             AGE
#> 2      <NA>     2       25-34             AGE             AGE
#> 3      <NA>     3       35-44             AGE             AGE
#> 4      <NA>     4         45+             AGE             AGE
#> 5       SEX     1      Female             SEX             SEX
#> 6      <NA>     2        Male             SEX             SEX
#> 7   SEGMENT     1           A         SEGMENT         SEGMENT
#> 8      <NA>     2           B         SEGMENT         SEGMENT
#> 9      <NA>     3           C         SEGMENT         SEGMENT
#> 10     <NA>     4           D         SEGMENT         SEGMENT
#> 11     <NA>     5           E         SEGMENT         SEGMENT

Answer 4

This replaces the "" with missing and then fixes the variable named Variable:

df %>% 
  dplyr::mutate_all(list(~na_if(.,""))) %>% 
  tidyr::fill(Variable, .direction = "down")

Answer 5

Using data.table and a for loop:

library(data.table)
DT <- as.data.table(df)

DT[, Variable_new := Variable[1]]

for (i in 2:nrow(DT)) {
  DT[i, Variable_new := fifelse(DT[i, Variable] == '', DT[i-1, Variable_new], DT[i, Variable])]
}

> DT
    Variable Value Description Variable_new
 1:      AGE     1       18-24          AGE
 2:              2       25-34          AGE
 3:              3       35-44          AGE
 4:              4         45+          AGE
 5:      SEX     1      Female          SEX
 6:              2        Male          SEX
 7:  SEGMENT     1           A      SEGMENT
 8:              2           B      SEGMENT
 9:              3           C      SEGMENT
10:              4           D      SEGMENT
11:              5           E      SEGMENT